Let $X$ be a compact topological space. Suppose that for any $x, y \in X$ with
$x \neq y$, there exist open sets $U_x$ and $U_y$ containing $x$ and $y$, respectively, such that $$ U_x \cup U_y = X\quad \text{and}\quad U_x \cap U_y = \varnothing.$$
Let $V \subseteq X$ be an open set. Let $x \in V$ . Show that there exists a set $U$ which is both open and closed and $x \in U \subseteq V$.
My Try:
$\forall \;x\neq y\quad U_x \cap U_y = \varnothing$, $X$ is Hausdorff hence a $T_1$-space. $X$ is compact Hausdorff hence normal subsequently normal $T_1$-space i.e. $T_4$-space. As a consequence of $T_1$-space, $\{x\}$ is closed in $X$.
Given $V$ is open and $x\in V$. So $\bbox[5px,border:1px solid red]{\text{there exists open set $U$ such that $x \in U \subseteq V.$}}$ $U^c$ is closed. By normality of $X$ there exists disjoint open sets $W_1$ and $W_2$ such that
$$\{x\}\subseteq W_1\subseteq U \;\text{and}\; U^c\subseteq W_2\implies W_2^c\subseteq U\tag 1$$
$$W_2^c\subseteq U\implies X=W_2\cup W_2^c\subseteq U \cup W_2 \tag 2$$
Claim: $U\cap W_2=\varnothing$.
For suppose $x\in U\cap W_2$ then,
$$x\in U \;\text{and}\; x\in W_2\implies\bbox[5px,border:1px solid red]{{ x\not\in U^c \;\text{or}\; x\not\in W^c_2\implies x\not\in U^c\cup W^c_2}}$$
Also from $(1)$ and $(2)$ we see that $$U^c\cup W^c_2\subseteq U\cup W_2\implies x\not\in U^c\cup W^c_2\subseteq U\cup W_2$$
a contradiction since $x\in U\cap W_2$ but $x\not\in U \cup W_2$. Hence the claim subsequently $U$ is both open and closed such that $x \in U \subseteq V$.
Is there anything incorrect or missing in my proof ? Are there any other alternative proofs?
Update: I figured the two highlighted portions are the incorrect parts of the proof. Thanks to @Thomas Andrews and @Hagen von Eitzen for clarifying my doubts and for their answers.
Best Answer
We are given (with improved notation) that for $x,y\in X$ with $x\ne y$, there exist open sets $U_{(x,y)}$ and $U_{(y,x)}$ such that $$x\in U_{(x,y)},\quad y\in U_{(y,x)},\quad U_{(x,y)}\cup U_{(y,x)} =X,\quad U_{(x,y)}\cap U_{(y,x)} =\emptyset. $$ Now assume $x\in X$ and $V\ni x$ is open. Then $V$ together with all $U_{(y,x)}$, $y\in V^\complement$ form an open cover of $X$. Pick a finite sub-cover consisting of $V$ and some $U_{(y_i,x)}$, $i=1,2,\ldots, n$. Let $$U=\bigcap_{i=1}^nU_{(x,y_i)}.$$ Then $U$ is a finite intersection if clopen sets, hence clopen. Clearly $x\in U$. And as the $U_{(y_i,x)}$ cover $V^\complement$, it follows that $U\subseteq V$.